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::::By the way, why don't you have a user page?
::::By the way, why don't you have a user page?
::::[[User:HOOTmag|HOOTmag]] ([[User talk:HOOTmag|talk]]) 08:56, 24 June 2010 (UTC)
::::[[User:HOOTmag|HOOTmag]] ([[User talk:HOOTmag|talk]]) 08:56, 24 June 2010 (UTC)
:::::You're welcome; I'm glad it was helpful.
:::::I've been using Wikipedia as a reference, mostly for math, and have enjoyed reading through the reference desks for quite a while. I have only occasionally participated, but I may well get an account eventually to do more editing and commenting. (Right now, I think I'm too busy working on a paper and my thesis.) I'll only be at this IP address for about a month more.
:::::[[Special:Contributions/98.235.80.144|98.235.80.144]] ([[User talk:98.235.80.144#top|talk]]) 20:07, 24 June 2010 (UTC)

Revision as of 20:07, 24 June 2010

Hello,

Do you agree with the following?

  • (Taken from the article Cauchy's integral theorem): For every function f which is holomorphic in a simply connected domain D, where F is the anti-derivative of f, the integral of f between a and b along a continuously differentiable path (in D), is: .
  • When F is the logarithmic function, then choosing a different domain may only change the value of and of , but it can't change the value of , since F (as an anti-derivative) must be continuous.

Looking forward to your response, HOOTmag (talk) 21:12, 23 June 2010 (UTC)[reply]

HOOTmag, the first statement is certainly correct; but I disagree with the second statement.
You are correct that for any fixed choice of domain, there are many antiderivatives F (for any choice of +C), but that F(b)-F(a) will always have the same value. But for different choices of domain, we can get different values for F(b)-F(a), because we may take a different path to get from a to b.
The problem with saying the antiderivative is F(z)=ln(z) is that the complex logarithm is a multivalued function. (The solution is to specify a branch cut.) You can see that -πi, πi, 3πi, 5πi, and infinitely many other choices could all be ln(-1), since .
Notice that if our domain D has a hole in it around z=0, then there is no continuous antiderivative of 1/z on D. It's like there's an imaginary spiral staircase around z=0. (See the picture at Branch point#Complex logarithm-- this picture is probably more helpful than anything I'm writing.) If you go around one way, you're off by 2πi; if you go around the other way, you're off by -2πi. You can take any path you want and find a continuous antiderivative until you've looped around and hit the same point twice-- the values don't match, so it can't be continuous.
So, if you pick a domain D that goes counterclockwise a half turn from -2 to 6, you travel one half turn up the staircase, which gives the +πi term. The continuity of F forces us to get this result, since we've gone a half turn counterclockwise. Similarly, if the domain D goes a half turn clockwise from -2 to 6, you travel one half turn down the staircase, which gives us -πi. We can choose domains D or branch cuts of ln(z) to get any of the other values-- 3πi, 5πi, etc. -- by spiraling around the origin on our way from -2 to 6.
98.235.80.144 (talk) 22:03, 23 June 2010 (UTC)[reply]
  1. Is the following correct? There is a continuous function F, satisfying exp(F(z))=z for every z in the domain of F, and there are a,b in this domain such that Im(F(b)-F(a))>2π.
  2. Is the following correct? There is a continuous function F, satisfying exp(F(z))=z for every z in the domain of F, so that for every integer n there is a number z such that F(z) ≠ ln(z)+2πin (ln being the principle logarithmic branch).
  3. If you think there is such a continuous function (as described in #1 or in #2), then could you describe it explicitly, using elementary operations/functions/constants only, like the principle logarithmic branch, π, i, and the like (piecewise-defined functions being permitted), i.e. without non-elementary tools as integrals and the like?
HOOTmag (talk) 22:41, 23 June 2010 (UTC)[reply]
Yes, there is such a function. Let the domain D be the set of points . (This is just a one-dimensional, counterclockwise spiral that crosses through z=1, z=2, z=3,...)
Then F(z) = ln(t) + 2πit satisfies your conditions. You can see that exp(F(z))=z, and that F(z) is continuous. We have F(1)=2πi, F(2)=ln(2)+4πi, F(3)=ln(3)+6πi, etc. If you try to define a complex logarithm G(z) on this domain that takes the principal values G(1)=0, G(2)=ln(2), G(3)=ln(3), etc., then it will necessarily be discontinuous.
If you want an open set for the domain, let D' be . This is just the complement of the domain D above, minus z=0. On D', we have F(z) = ln(t) + 2πis that satisfies all of your conditions above; it is just the complex logarithm with the spiral D as the branch cut.
98.235.80.144 (talk) 02:34, 24 June 2010 (UTC)[reply]
Thank you a lot for your instructive explanation, and for your patience as well. Now I know this point was my principle error which caused all of my other errors. I've always thought that every continuous function F, satisfying exp(F(z))=z for every z in the domain of F, has a constant integer n satisfying F(z) = ln(z)+2πin for every z in the domain of F (ln being the principle logarithmic branch), because I couldn't imagine that excluding a spiral (as you've well described) out of the plane - gives a simply connected domain on which one can define a continuous logarithmic branch such that refutes my wrong thought about the pseudo-constant integer of continuous logarithmic branches. Again, thank you much.
By the way, why don't you have a user page?
HOOTmag (talk) 08:56, 24 June 2010 (UTC)[reply]
You're welcome; I'm glad it was helpful.
I've been using Wikipedia as a reference, mostly for math, and have enjoyed reading through the reference desks for quite a while. I have only occasionally participated, but I may well get an account eventually to do more editing and commenting. (Right now, I think I'm too busy working on a paper and my thesis.) I'll only be at this IP address for about a month more.
98.235.80.144 (talk) 20:07, 24 June 2010 (UTC)[reply]